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Question Number 187050    Answers: 0   Comments: 1

∫^(𝛑/4) _0 ((cos^(−1) x + cos x)/(Ln x)) dx =??

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\:\frac{\boldsymbol{{cos}}^{−\mathrm{1}} \boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\:\boldsymbol{{x}}}{\boldsymbol{{Ln}}\:\boldsymbol{{x}}}\:\:\boldsymbol{{dx}}\:=??\:\:\: \\ $$$$ \\ $$

Question Number 187046    Answers: 1   Comments: 0

find the general solution for the following system of equations ((dx_1 /dt))=2x_1 +2x_2 ((dx_2 /dt))=x_1 +3x_2

$${find}\:\:\:{the}\:{general}\:\:{solution}\:{for}\:\:{the}\:{following} \\ $$$${system}\:{of}\:{equations} \\ $$$$\left(\frac{{dx}_{\mathrm{1}} }{{dt}}\right)=\mathrm{2}{x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} \\ $$$$\left(\frac{{dx}_{\mathrm{2}} }{{dt}}\right)={x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} \\ $$$$ \\ $$

Question Number 187029    Answers: 1   Comments: 0

Question Number 187027    Answers: 1   Comments: 0

Question Number 187025    Answers: 1   Comments: 0

Question Number 187044    Answers: 0   Comments: 0

f(x,y)={(x+y),0<x<1, 0<y<1 0, otherwise find: (1) σxy (2) p(0<x<((1/4))∣((2/5))>y>((1/3))) note: f(x,y) is the joint PDF

$${f}\left({x},{y}\right)=\left\{\left({x}+{y}\right),\mathrm{0}<{x}<\mathrm{1},\:\mathrm{0}<{y}<\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0},\:{otherwise} \\ $$$${find}: \\ $$$$\left(\mathrm{1}\right)\:\sigma{xy} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:{p}\left(\mathrm{0}<{x}<\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\mid\left(\frac{\mathrm{2}}{\mathrm{5}}\right)>{y}>\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$ \\ $$$${note}:\:{f}\left({x},{y}\right)\:{is}\:{the}\:{joint}\:{PDF} \\ $$

Question Number 187020    Answers: 0   Comments: 1

(a/x)=(b/y)=(c/z)=(1/3) ,a−2b+c=2 and 2y−3z=1 x=? how is solution

$$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:,{a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:\mathrm{2}{y}−\mathrm{3}{z}=\mathrm{1}\:\:\:\:{x}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 187011    Answers: 0   Comments: 0

a∈C determinant (((a 1 … 1)),((1 ⋱ ⋱ ⋮)),((⋮ ⋱ ⋱ 1)),((1 … 1 a)))=?

$$\mathrm{a}\in\mathbb{C} \\ $$$$\begin{vmatrix}{\mathrm{a}\:\:\:\mathrm{1}\:\:\:\:\ldots\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\ddots\:\:\ddots\:\:\vdots}\\{\vdots\:\:\ddots\:\:\ddots\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\ldots\:\:\:\:\mathrm{1}\:\:\:\mathrm{a}}\end{vmatrix}=? \\ $$

Question Number 187010    Answers: 0   Comments: 0

a,b∈C determinant (((a+b a 0 … 0)),(( b a+b ⋱ ⋱ ⋮)),(( 0 ⋱ ⋱ ⋱ 0)),(( ⋮ ⋱ ⋱ ⋱ a)),(( 0 … 0 b a+b)))=?

$$\mathrm{a},\mathrm{b}\in\mathbb{C} \\ $$$$\:\begin{vmatrix}{\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\vdots}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\vdots\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}}\end{vmatrix}=? \\ $$

Question Number 187008    Answers: 2   Comments: 0

(a/3)=(b/4)=(c/5) 3a+c=42 b=? how is solution

$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 187007    Answers: 0   Comments: 2

prove that 7^(k+1) ∣ 13^(7k) + 1 , ∀k∈N

$${prove}\:\:{that} \\ $$$$\mathrm{7}^{{k}+\mathrm{1}} \mid\:\mathrm{13}^{\mathrm{7}{k}} \:+\:\mathrm{1}\:,\:\forall{k}\in\mathbb{N} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 186999    Answers: 0   Comments: 0

Question Number 186998    Answers: 1   Comments: 1

Question Number 186996    Answers: 1   Comments: 0

Question Number 186989    Answers: 0   Comments: 0

Question Number 186983    Answers: 3   Comments: 0

Question Number 186960    Answers: 1   Comments: 0

Question Number 186958    Answers: 0   Comments: 0

Question Number 186952    Answers: 2   Comments: 0

Question Number 186951    Answers: 2   Comments: 0

Question Number 186950    Answers: 0   Comments: 0

Question Number 186941    Answers: 0   Comments: 1

((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b ) prove that

$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$

Question Number 186929    Answers: 3   Comments: 0

If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?

$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$

Question Number 186926    Answers: 1   Comments: 1

Question Number 186924    Answers: 3   Comments: 0

Question Number 186923    Answers: 0   Comments: 0

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